Sec(6) theetha=tan(6)thetha+ 3tan(2)thetha(sec(2)thetha+1

 We have to prove that :- 

{tex} \Rightarrow {\sec ^6}\theta = {\tan ^6}\theta +3{\tan ^2}\theta {\sec ^2}\theta + 1{/tex}

{tex} \Rightarrow {\sec ^6}\theta - {\tan ^6}\theta = 3{\tan ^2}\theta {\sec ^2}\theta + 1{/tex}

Now, LHS

 {tex} = {\sec ^6}\theta - {\tan ^6}\theta {/tex}
{tex} = {\left( {{{\sec }^2}\theta } \right)^3} - {({\tan ^2}\theta )^3}{/tex}

{tex}= \left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right)\left[ {{{({{\sec }^2}\theta )}^2} + {{\sec }^2}\theta {{\tan }^2}\theta + {{({{\tan }^2}\theta )}^2}} \right]{/tex}{Since, a³ - b³ = (a - b )(a² - ab + b² )}
{tex} = 1\left[ {{{\sec }^4}\theta + {{\sec }^2}\theta {{\tan }^2}\theta + {{\tan }^4}\theta } \right]{/tex} {tex}\left[ {\because {{\sec }^2}\theta - {{\tan }^2}\theta = 1} \right]{/tex}
{tex} = {\sec ^4}\theta + {\tan ^4}\theta + {\sec ^2}\theta {\tan ^2}\theta {/tex}
{tex} = {({\sec ^2}\theta )^2} + {({\tan ^2}\theta )^2} + {\sec ^2}\theta {\tan ^2}\theta {/tex}
Adding and subtracting {tex}2{\sec ^2}\theta {\tan ^2}\theta {/tex}
{tex} = {({\sec ^2}\theta )^2} + {({\tan ^2}\theta )^2} - 2{\sec ^2}\theta {\tan ^2}\theta + 2{\sec ^2}\theta {\tan^2}\theta+ {\sec ^2}\theta {\tan ^2}\theta {/tex}
{tex} = {({\sec ^2}\theta - {\tan ^2}\theta )^2} + 3{\sec ^2}\theta {\tan ^2}\theta {/tex} {tex}\left[ {\because {{(a - b)}^2} = {a^2} + {b^2} - 2ab} \right]{/tex}
{tex} = 1 + 3{\sec ^2}\theta {\tan ^2}\theta {/tex} {tex}\left[ {\because {{\sec }^2}\theta - {{\tan }^2}\theta = 1} \right]{/tex}
= RHS
Hence proved